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Difficulty: MediumLoad Distribution and Balancing

A uniform 1212-foot diving board weighing 5050 pounds is attached to a rear tie-down anchor at one end. The board rests on a fulcrum support located exactly 33 feet from the rear anchor. A 100100-pound swimmer stands at the extreme opposite end of the board. What is the required downward holding force (tension) at the rear tie-down anchor to keep the board horizontally balanced?

  1. A
    150150 pounds
  2. B
    300300 pounds
  3. 350350 poundsAnswer
  4. D
    500500 pounds

Answer

350350 pounds
To find the balancing force, we must ensure all rotational forces (torques) around the fulcrum cancel out. The swimmer (100100 lbs at a 99-foot distance from the fulcrum) creates 900900 lb-ft of torque. The board's center of gravity (5050 lbs at a 33-foot distance from the fulcrum) creates 150150 lb-ft of torque. The total clockwise torque is 10501050 lb-ft. Because the rear anchor is 33 feet from the fulcrum on the opposite side, it must pull down with a force of 1050÷3=3501050 \div 3 = 350 pounds to maintain equilibrium.

Step-by-Step Solution

1
Identify the pivot point (fulcrum) and calculate the distances of all forces from this point.
The fulcrum is at 33 ft. The rear anchor is 33 ft away. The board's center of gravity is at 66 ft (33 ft from fulcrum). The swimmer is at 1212 ft (99 ft from fulcrum).
Torque is calculated by multiplying force by the distance to the fulcrum, not to the edge of the object.
2
Calculate the total clockwise torque caused by the board and the swimmer.
Board torque: 50 lbs×3 ft=150 lb-ft50 \text{ lbs} \times 3 \text{ ft} = 150 \text{ lb-ft}. Swimmer torque: 100 lbs×9 ft=900 lb-ft100 \text{ lbs} \times 9 \text{ ft} = 900 \text{ lb-ft}. Total clockwise torque = 1050 lb-ft1050 \text{ lb-ft}.
These forces push down on the right side of the fulcrum, acting to rotate the front of the diving board downward.
3
Determine the downward force required at the rear anchor to balance the system.
The rear anchor must provide an equal counter-clockwise torque. Force×3 ft=1050 lb-ft\text{Force} \times 3 \text{ ft} = 1050 \text{ lb-ft}. Force=10503=350 lbs\text{Force} = \frac{1050}{3} = 350 \text{ lbs}.
For static equilibrium, the opposing torques around the fulcrum must be equal.

Key Concept

Static Equilibrium and Torque
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Topics

Load Distribution and Balancing

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